Problem: What is the extraneous solution to these equations? $\dfrac{x^2 + 2x}{x - 1} = \dfrac{7x - 4}{x - 1}$
Explanation: Multiply both sides by $x - 1$ $ \dfrac{x^2 + 2x}{x - 1} (x - 1) = \dfrac{7x - 4}{x - 1} (x - 1)$ $ x^2 + 2x = 7x - 4$ Subtract $7x - 4$ from both sides: $ x^2 + 2x - (7x - 4) = 7x - 4 - (7x - 4)$ $ x^2 + 2x - 7x + 4 = 0$ $ x^2 - 5x + 4 = 0$ Factor the expression: $ (x - 4)(x - 1) = 0$ Therefore $x = 4$ or $x = 1$ At $x = 1$ , the denominator of the original expression is 0. Since the expression is undefined at $x = 1$, it is an extraneous solution.